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12 A 4.8 (2 m 1.92 (2 161, 80 15/02A 402 130 wwmm 20 -1452 Fig. 6 Fig.7 Q10) The average power supplied by the dependent source of Fig. 6 can be determined as c) 96 W d) 192 W c) 48W a) 5W b) 24 W Q11) The Vh for the the circuit shown in Fig. 7 ar a)-(220 Y 6.54,

12 A 4.8 (2 m 1.92 (2 161, 80 15/02A 402 130 wwmm 20 -1452 Fig. 6 Fig.7 Q10) The average power supplied by the dependent source of Fig. 6 can be determined as c) 96 W d) 192 W c) 48W a) 5W b) 24 W Q11) The Vh for the the circuit shown in Fig. 7 ar a)-(220 Y 6.54,

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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circuits, pleaseeeee solve questionnn10
The switch in Fig. 1 has been closed for long time. It opens at t=0. Please refer to the circuit of Fig. 1 for the following questions (Q1, and Q2) Q1) The time constant t can be found as: a) 6.67 s b) 0.3 s c) 10 s d) 0.1 s Q2) The current i(t) at t= 1m s is: a) 2.02 A b) 6 A c) 4.02 A a) 1.23 cos(10t +30°) V d) 2.25 cos(10t-53.6%) V e) 0.15 s d) 5.98 A e) 4 A 2cos101 V b) 1.23 cos(10t-30°) V e) 1.79 cos(10t -26.57°) V 20 400 40 www HE Refer to the circuit of Fig. 2 for the following 3 questions (Q3, Q4 and Q5) Q3) By using superposition technique, the contribution of the 2cos10t voltage source to the value of vi(t) is: Q5) The value of the inductance of the j2 2 impedance is: a) 0.2 H b) 10 H c) 20 H d)1.6 H e) 16 H Q7) The current in(t) of Fig. 4 can be found as (mA): a) 12.5cos(500t - 0.107°) d) 12.5 cos(500t + 89.9°) pa Q6) Referring to the circuit of Fig. 3, Zin can be determined as: a)22+j6Ω b)18+j6Ω c) 22-j6 Ω d) 18-j62 e)-18+j6 22 Q4) By applying KCL to the node v/(t), the value of the voltage labeled v/(t) is (V): a) 2.86 cos(10t +77.9°) b) 2.86 cos(10t-77.9°) d) 4.1 cos(10t-62.3°) c) 4.1 cos(10t +62.3°) f) 3.92 cos(10t-77.9°) e) 3.92 cos(10t +77.9°) 5923 5 cos 10rv b) 12.5cos(500t+ 0.107°) e) 12.5 cos(500t+ 0.205°) 20:2 -ww c) 2.25 cos(10t +53.6°) V f) 1.79 cos(10t+26.57°) V 10.0 ww 7 1.5 H m 2 -15.02 34 - 1=0 Q9) The complex power absorbed by voltage source is (VA) b)-751.3-j457. c)-823.5+j294.1 a) -823.5-j294.1 d) -751.3+j457.7 e) 751.3-j457.7 Fig. 1 102 cos5001 V Fig. 2 Fig. 3 Refer to the circuit of Fig. 5 for the following 2 questions (Q8, and Q9) Q8) The current through the-j10 2 can be found as (Arms) a) 8.75/19.65* b) 8.75-19.65* c) 10.25/90* d) 10.25Z-90° e) 202-53.26 f) 20253.26 c) 12.5cos(500t - 89.9°) f) 12.5 cos(500t - 0.205°) 10Ω (1) 6A 100/0° V ms 10042 www 0.2 i Fig. 4 –ΠΟΥ Fig. 5 31 2002 0.3mH 200 ww 1002
Transcribed Image Text:The switch in Fig. 1 has been closed for long time. It opens at t=0. Please refer to the circuit of Fig. 1 for the following questions (Q1, and Q2) Q1) The time constant t can be found as: a) 6.67 s b) 0.3 s c) 10 s d) 0.1 s Q2) The current i(t) at t= 1m s is: a) 2.02 A b) 6 A c) 4.02 A a) 1.23 cos(10t +30°) V d) 2.25 cos(10t-53.6%) V e) 0.15 s d) 5.98 A e) 4 A 2cos101 V b) 1.23 cos(10t-30°) V e) 1.79 cos(10t -26.57°) V 20 400 40 www HE Refer to the circuit of Fig. 2 for the following 3 questions (Q3, Q4 and Q5) Q3) By using superposition technique, the contribution of the 2cos10t voltage source to the value of vi(t) is: Q5) The value of the inductance of the j2 2 impedance is: a) 0.2 H b) 10 H c) 20 H d)1.6 H e) 16 H Q7) The current in(t) of Fig. 4 can be found as (mA): a) 12.5cos(500t - 0.107°) d) 12.5 cos(500t + 89.9°) pa Q6) Referring to the circuit of Fig. 3, Zin can be determined as: a)22+j6Ω b)18+j6Ω c) 22-j6 Ω d) 18-j62 e)-18+j6 22 Q4) By applying KCL to the node v/(t), the value of the voltage labeled v/(t) is (V): a) 2.86 cos(10t +77.9°) b) 2.86 cos(10t-77.9°) d) 4.1 cos(10t-62.3°) c) 4.1 cos(10t +62.3°) f) 3.92 cos(10t-77.9°) e) 3.92 cos(10t +77.9°) 5923 5 cos 10rv b) 12.5cos(500t+ 0.107°) e) 12.5 cos(500t+ 0.205°) 20:2 -ww c) 2.25 cos(10t +53.6°) V f) 1.79 cos(10t+26.57°) V 10.0 ww 7 1.5 H m 2 -15.02 34 - 1=0 Q9) The complex power absorbed by voltage source is (VA) b)-751.3-j457. c)-823.5+j294.1 a) -823.5-j294.1 d) -751.3+j457.7 e) 751.3-j457.7 Fig. 1 102 cos5001 V Fig. 2 Fig. 3 Refer to the circuit of Fig. 5 for the following 2 questions (Q8, and Q9) Q8) The current through the-j10 2 can be found as (Arms) a) 8.75/19.65* b) 8.75-19.65* c) 10.25/90* d) 10.25Z-90° e) 202-53.26 f) 20253.26 c) 12.5cos(500t - 89.9°) f) 12.5 cos(500t - 0.205°) 10Ω (1) 6A 100/0° V ms 10042 www 0.2 i Fig. 4 –ΠΟΥ Fig. 5 31 2002 0.3mH 200 ww 1002
2A (4) le 4.8 2 am Q12) Neper frequency a is a) 4 rad/s b) 16 rad/s c) 25 rad/s d) 2 rad/s e) 5 rad/s Q13) Resonant frequency o is a) 4 rad/s b) 5 rad/s c) 4 rad/s d) 16 rad/s e) 25 rad/s Q14) The voltage v (t=1s) can be found as a) 10.61 V b) 14.61 V c) 16.61 V d) 8.61 V e) 12.61 V a) R=0.202, C=0.2 F d) R=102, C=1F m 1.92 02 34 L m L Fig. 6 Fig.7 Q10) The average power supplied by the dependent source of Fig. 6 can be determined as b) 24 W d) 192 W a) 5W c) 96 W c) 48W Q11) The Va for the the circuit shown in Fig. 7 as seen from the terminal a-b can be found as: a)-/220 V d)-110 V )-/165 V c)-j55 V b)-/330 V Please refer to the circuit of Fig. 8 for the following questions (Q12, Q13 and Q14). Assuming it(0)-8 A, and V.(0)-40 V. T 1.61, 80 30 34 L m Z Fig. 9 Q15) For the circuit shown in Fig. 9, the equivalent inductance Leq is: a) 1/2 L d) 5/8 L e) 4/7 L b) 4/9 L c) 7/4 L Q19) For the circuit shown in Fig. 11, the value of C needed to make the response underdamped with unity damping factor (a = 1) is: 1520 A e) 24 V e) 15 V SH 6 MA 24 V + 14 3k0z www 402 -NN 20 +49 ΣΕΩΣ ΔΩΣ t=0 3kQ2 Fig. 8 2mF HF 2102 ww J30 400 Q16) For the circuit shown in Fig. 10, the energy stored in the 4 mF capacitor under de conditions is: a) 32 mJ d) 8 mJ e) 16 mJ b) 128 mJ c) 256 mJ Q17) If v(t)=15 cos(1000t+66°) V and i(t)=2cos(1000t+450°) A, then v(t) leads i(t) by a) 156° b)-24° c) 204° d) 24° e) 66° Q18) Assuming that the input impedance is given as Zin= 1+j1 02 and co-1 rad/s, then the input admittance can be represented as the parallel combination of: b) R=0.2 02, L=0.2 H e) R=202, L=2H c) R=10, L=1H Fig. 10 + 100 0.5H C= 10 mF SKΩ Σ 4 m² = a) 40 mF b) 15 mF c) 26 mF d) 2.5 mF e) 7.5 mF The switch in Fig. 12 has been in position A for long time. At t=0, the switch moves to position B. Please refer to the circuit of Fig. 12 for the following questions (Q20, Q21 and Q22) Fig. 11 Q20) v(0) can be found as: a)30 V b) 12.5 V c) 9V d) 15 V Q21) v(co) can be found as: a) 24 V b)30 V c) 12.5 V d) 9 V Q22) The voltage v(t) at t = 1 s is: a) 20.9 V b)24.9 V c) 30 V d) 39.1 V e) 27.97 V Fig. 12 12.5mFv 04 B 4kQ co www 9.51, P0.5 mF b 410 30 V
Transcribed Image Text:2A (4) le 4.8 2 am Q12) Neper frequency a is a) 4 rad/s b) 16 rad/s c) 25 rad/s d) 2 rad/s e) 5 rad/s Q13) Resonant frequency o is a) 4 rad/s b) 5 rad/s c) 4 rad/s d) 16 rad/s e) 25 rad/s Q14) The voltage v (t=1s) can be found as a) 10.61 V b) 14.61 V c) 16.61 V d) 8.61 V e) 12.61 V a) R=0.202, C=0.2 F d) R=102, C=1F m 1.92 02 34 L m L Fig. 6 Fig.7 Q10) The average power supplied by the dependent source of Fig. 6 can be determined as b) 24 W d) 192 W a) 5W c) 96 W c) 48W Q11) The Va for the the circuit shown in Fig. 7 as seen from the terminal a-b can be found as: a)-/220 V d)-110 V )-/165 V c)-j55 V b)-/330 V Please refer to the circuit of Fig. 8 for the following questions (Q12, Q13 and Q14). Assuming it(0)-8 A, and V.(0)-40 V. T 1.61, 80 30 34 L m Z Fig. 9 Q15) For the circuit shown in Fig. 9, the equivalent inductance Leq is: a) 1/2 L d) 5/8 L e) 4/7 L b) 4/9 L c) 7/4 L Q19) For the circuit shown in Fig. 11, the value of C needed to make the response underdamped with unity damping factor (a = 1) is: 1520 A e) 24 V e) 15 V SH 6 MA 24 V + 14 3k0z www 402 -NN 20 +49 ΣΕΩΣ ΔΩΣ t=0 3kQ2 Fig. 8 2mF HF 2102 ww J30 400 Q16) For the circuit shown in Fig. 10, the energy stored in the 4 mF capacitor under de conditions is: a) 32 mJ d) 8 mJ e) 16 mJ b) 128 mJ c) 256 mJ Q17) If v(t)=15 cos(1000t+66°) V and i(t)=2cos(1000t+450°) A, then v(t) leads i(t) by a) 156° b)-24° c) 204° d) 24° e) 66° Q18) Assuming that the input impedance is given as Zin= 1+j1 02 and co-1 rad/s, then the input admittance can be represented as the parallel combination of: b) R=0.2 02, L=0.2 H e) R=202, L=2H c) R=10, L=1H Fig. 10 + 100 0.5H C= 10 mF SKΩ Σ 4 m² = a) 40 mF b) 15 mF c) 26 mF d) 2.5 mF e) 7.5 mF The switch in Fig. 12 has been in position A for long time. At t=0, the switch moves to position B. Please refer to the circuit of Fig. 12 for the following questions (Q20, Q21 and Q22) Fig. 11 Q20) v(0) can be found as: a)30 V b) 12.5 V c) 9V d) 15 V Q21) v(co) can be found as: a) 24 V b)30 V c) 12.5 V d) 9 V Q22) The voltage v(t) at t = 1 s is: a) 20.9 V b)24.9 V c) 30 V d) 39.1 V e) 27.97 V Fig. 12 12.5mFv 04 B 4kQ co www 9.51, P0.5 mF b 410 30 V
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